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Fluid Flow Losses in Pneumatic Design: Balloon Filling Struggles

Explore the challenges of fluid flow losses in pneumatic systems when filling a balloon using ¼ inch ID tubes, 7 feet long, and a flow rate of 35 liters per minute. Discover the implications of Bernoulli's equation, Pitot tubes, Venturi effect, and more. Learn about the impact of pressure, density, and velocity on centrifugal pumps and balloons, and understand the significance of Euler curvature in fluid mechanics. Delve into the complexities of fluid dynamics and energy transfer in a pneumatic system. Additional resources and practical applications of fluid mechanics and thermodynamics in design and manufacturing are discussed.

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Fluid Flow Losses in Pneumatic Design: Balloon Filling Struggles

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  1. 2.007 –Design and Manufacturing IPneumatics: and fluid flow more generally Dan Frey

  2. What are the Losses in the Tube When You Fill the Balloon? • ¼ inch ID tube • 7 feet long • 35 liters per minute flow rate

  3. What are the Losses in the Tube When You Fill the Balloon? • 35 liters per minute flow rate • ¼ inch ID tube • 7 feet long • About 0.3 psi

  4. Flow in a Pipe • This is called a “Moody diagram”

  5. What Losses are Worse, the Ones in the Table or the Ones in your Robot? • 35 liters per minute flow rate • 1/8 inch ID tube • 7 inches long • About 0.7 psi

  6. Flow in a Pipe • This is called a “Moody diagram”

  7. Euler and Streamline Curvature • Pressure gradient perpendicular to streamline

  8. Euler and Streamline Curvature • So, just about 4 in height diff between the edge and the center

  9. What Does this Tell You About Centrifugal Pumps and Balloons? • A BP-05 at 262 rpm with 5 cm radius spinning water can produce a pressure rise of 4 in of water • You need about 1 psi (~30 inches water, almost 10X) • You are spinning air, not water (about 1000X) • So pressure/density is a 10,000X deficit • Pressure/density rise goes like w2R2 (tip velocity squared) • To get a 10,000X increase, you need 100X in wR (tip velocity) • If you stay at 5cm radius, you need a motor that runs at 26,000 rpm at your chosen voltage (also 1psi*35*liter/min=4Watts) • Or, with a slower motor (e.g., BP-05, take off the 40:1 gear box and increase the radius or voltage by a factor of 5) Also consider efficiency BUT, the effect in the water experiment is only about HALF the story

  10. Bernoulli’s Equation • Inertia forces >> viscous forces • Applies along a streamline • Constant density • Steady flow (this form of the eqn) http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-28/

  11. Bernoulli’s Equation Static pressure Dynamic pressure (kinetic energy per unit volume) Gravitational potential energy per unit volume http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-28/

  12. Classic Application of Bernoulli:Water Flow from a Pressurized Tank • Fill the 2 liter bottle partially with water • Let’s pump the bottle to 30psi gauge • When I release the clamp, how high will the water go?

  13. Here is my approach • Go check out “water height problem v5.mcd” on Stellar

  14. Implication of Bernoulli:The Venturi Effect • Reducing cross sectional area causes • Velocity to rise (inversely proportional to area) • Dynamic pressure (rises as velocity squared) • So, static pressure drops • Can be used to measure flow velocity Pitot tube

  15. How an Air Speed Indicator Works • Total pressure indicated by ram air • Static pressure collected also • Difference measured and amplified by diaphragm • Mechanism converts into dial indication

  16. Other Implication of Bernoulli:Diffusers in Compressors • In the spinning water problem, pressure rise was • But we also imparted velocity to the flow, and we could recover some static pressure rise from that Dixon, Fluid mechanics and thermodynamics of turbomachinery http://library.books24x7.com.libproxy.mit.edu/book/id_36368/

  17. Viscous Forces in Flows • There is some velocity distribution • At the surface, the shear stress is y V(y) v(y)=m(dV/dy)

  18. Plane Poiseuille Flow • If viscous forces >> inertia forces • Velocity distribution is parabolic V(y) y h x

  19. Poiseuille and Connecting • Prof. Gossard’s connector • A very simple thing I can analyze • Assume 0.001inch slip fit all along intersection

  20. Plane Poiseuille Flow • h=0.001 inch • dp=1psi • dx=7mm • m=3.5x10-7lb*sec/ft^2 • W=p*0.315inch • Q=0.12 liter/min y V(y) x So, quite a slow leak.

  21. Plane Poiseuille Flow:Check the Assumptions • Q=0.12 liter/min • h=0.001 inch • W=p*0.315inch • V= Q/(h*W) = 3m/s • Re=6.0 y V(y) x So, dominated by viscous forces.

  22. Plane Poiseuille Flow:Check the Forces Involved y • Force ≈ 1/20 lb x So, no trouble keeping it on.

  23. Pneumatic System The whole thing is ~ 0.2 kg Cylinder 50 gr Valve 25 gr Bottle, fitting, hose, etc, 50 gr What can I do with it? How much: force? energy? power?

  24. How much work does it take to fill a 2 liter bottle to 60 psi? 15psi 15psi First, how much air do we need? We assume constant temperature, so PV=const. Note: Atmospheric pressure is about 15psi (14.7psi) NOTE: We are talking about 60psi on the gauge, that’s about 75 psi absolute. So P1V1=P2V2 and since the ratio of pressures is 5:1, so is the ratio of volumes. It takes ~10 liters of air from atmospheric conditions. 15 psi 75psi 15psi 15psi

  25. F1 How much work does it take to fill a 2 liter bottle to 60 psi? 15psi A quick estimate. F2~4000N A=(10cm)^2 15psi The relationship is not linear really, we’ll think about that next. F2 15 psi 75psi 20cm 15psi x 20cm 100cm Assuming a linear change in force W=area under the curve = ½ * 4100N*80cm~1600J This is an OVER estimate 15psi

  26. F1 How much work does it take to fill a 2 liter bottle to 60 psi? 15psi P2=75psi A refined estimate. A=(10cm)^2 15psi F2 What is this, physically? 15 psi 75psi 10 liters. 15psi V 2 liters P1=15psi 10 liters 15psi

  27. Comparison • Assuming 60 psi • 5 lbf • 2 inch stroke • <0.1 sec • >11W • SY113-SMO Solenoid Actuated Valve • Maximum flow rate = 10 liters / minute • Displacement of cylinder 4cm^3 Pneumatic actuators in the kit > 2X more powerful even though they are lighter.

  28. Comparison • Assuming 60 psi • 5 lbf • 2 inch stroke • 0.1 sec • ~11W • Assuming 3V • 0.5 Nm • 30 rpm • 1.5W • 4W at 4.8V Pneumatic actuators in the kit roughly 2X more powerful even though they are lighter.

  29. Compressed Air: Benefits • Simple linear motion • High forces • Rapid motion • Potential for high power in a small, light package

  30. Compressed Air: Hazards • High forces • Rapid motion • Potential for high power Show movie

  31. Next Steps • Spring break next week • Lab open from 8:30AM to 4:30 PM Monday to Thursday • Lab closed Friday • HW#4 will be issued Tuesday after break • It will be short, like 2 hours • Due 1.5 weeks later on 7 April • Exam #2 will be Thursday 14 April

  32. Adiabatic and Isothermal Processes

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